Prove that ` sum_(r=1)^(n)(-1)^(r-1)(1+1/2+1/3+"....."+1/r)""^(n)C_(r) = 1/n`.

Explanation

`sum(-1)^(r-1).^(n)C_(r)(1/1+1/2+1/3+"...."+1/r)`
`=sum((-1)^(r-1).^(n)C_(r)underset(0)overset(1)int(1+x+x^(2)+"....."+x^(r-1))dx)`
`=sum(-1)^(r-1)..^(n)C_(r)underset(0)overset(1)int((1-x^(r))/(1-x))dx`
`= underset(0)overset(1)intunderset(r=1)overset(n)sum((-1)^(r-1)..^(n)C_(r)-(-1)^(r-1)..^(n)C_(r)x^(r))/(1-x)dx`
`=underset(0)overset(1) int(.^(n)C_(0)+(1-x)^(n) + (1-x)^(n))/(1-x)dx`
`= underset(0)overset(1)int(1-x)^(n-1)dx`
`= [(-(1-x)^(n-1))/(n-1)]_(0)^(1)`
`= 1/n`